# Prove $R\times S$ is not an integral domain

Let $$R$$ and $$S$$ be two commutative rings with unity. Prove that $$R\times S$$ is NOT an integral domain.

This is the best I could think of so far, please give me a push in the right direction and correct me.

It suffices to show that $$R\times S$$ has zero-divisors

Therefore, let $$a$$ be an element of $$R$$ and $$b$$ be an element of $$S$$, thus $$R\times S$$ has elements of the form $$(a,b)$$.

Consider the elements $$A=(a,0)$$ and $$B=(0,b)$$, such that $$a$$ does not equal zero and $$b$$ does not equal zero. Then $$AB = 0$$. Therefore $$R\times S$$ has zero divisors.

• Your argument is correct, but what is your question? – Crostul May 11 '15 at 18:06
• the question was proving that R+S is not an integral domain? I wanted to know if my proof and logic was correct – B ry May 11 '15 at 18:08
• Your argument is correct then. – jgon May 11 '15 at 18:09
• @Bry If you bothered to use the search function, you would have found confirmation at a question entitled "Prove R×R is NOT an integral domain." – rschwieb May 11 '15 at 19:04
• I used it and I guess I didn't realize RxR is the same as external direct product – B ry May 11 '15 at 19:25

It would help the clarity of your proof (and, on a fairly pedantic level, would make your proof more correct) if you chose specific non-zero elements $a,b$. In particular, we could take $a = 1_R$ and $b = 1_S$.
This way, we could always say that we're explicitly using the fact that a ring with unity has both a $1$ and a $0$.
• Indeed, $1_R$ and $1_S$ are the only elements which we know to exist and be $\ne 0$ in the given rings – Hagen von Eitzen May 11 '15 at 18:13
• Sorry to be pedantic, but I don't think $1 \not= 0$ is usually taken as one of the ring axioms. For once, Wikipedia bears me out: en.wikipedia.org/wiki/… en.wikipedia.org/wiki/Zero_ring – Rob Arthan May 11 '15 at 21:16